Problem: Simplify the following expression and state the condition under which the simplification is valid. You can assume that $y \neq 0$. $r = \dfrac{y}{7y - 42} \times \dfrac{3y - 18}{-5} $
Answer: When multiplying fractions, we multiply the numerators and the denominators. $r = \dfrac{ y \times (3y - 18) } { (7y - 42) \times -5 } $ $ r = \dfrac {y \times 3(y - 6)} {-5 \times 7(y - 6)} $ $ r = \dfrac{3y(y - 6)}{-35(y - 6)} $ We can cancel the $y - 6$ so long as $y - 6 \neq 0$ Therefore $y \neq 6$ $r = \dfrac{3y \cancel{(y - 6})}{-35 \cancel{(y - 6)}} = -\dfrac{3y}{35} = -\dfrac{3y}{35} $